CAT 2019 Quant Question Paper with Solution (Slot -2)

Let their salaries in 2010 be 6x, 5x and 7x respectively.

Also, let their salaries in 2015 be 3y, 4,y and 3y respectively

Given, $3y = 1.25 \times 6x$

Or y = 2.5x.

Therefore, salary of Rajesh in 2015 = $3y = 3 \times 2.5x=7.5x$

Percentage increase = $\left( \frac{7.5x-7x}{7x} \right)\times 100\approx 7%$

The expression will be real only if ${{\log }_{e}}\frac{4x-{{x}^{2}}}{3}\ge 0$

Or $\frac{4x-{{x}^{2}}}{3}\ge {{e}^{0}}$

$\Rightarrow \frac{4x-{{x}^{2}}}{3}\ge 1$

$\Rightarrow 4x-{{x}^{2}}\ge 3$

$\Rightarrow {{x}^{2}}-4x+3\le 0$

$\Rightarrow \left( x-1 \right)\left( x-3 \right)\le 0$

$1 \leq x \leq 3$

Let their scores after review be 11x, 10x, and 3x respectively.

Therefore, their scores before review was: (11x-6), 10x-6) and (3x-6) respectively.

Given, Rama's score was one-twelfth of the sum of the scores of Mohan and Anjali.

$\Rightarrow \left( 3x-6 \right)=\frac{1}{12}\left[ \left( 11x-6 \right)+\left( 10x-6 \right) \right]$

$\Rightarrow 12\left( 3x-6 \right)=21x-12$

$\Rightarrow 36x-72=21x-12$

$\Rightarrow 36x-21x=72-12=60$

$\Rightarrow x=4$

Now, Anjali’s score – Rama’s score = (11x-6)-(3x-6)=8x =32.

Shortcut:

Number of pairs = $\frac{number\ of\ factors\ 105}{2}$

$105=3\times 5\times 7$

Number of factors = $2\times 2\times 2=8$

Hence, required number of pairs =8/2 =4

Detailed Explanation:

${{\text{m}}^{2}}+105={{\text{n}}^{2}}$

$\Rightarrow {{n}^{2}}-{{m}^{2}}=105$

$\Rightarrow \left( n-m \right)\left( n+m \right)=105$

Since m and n are positive integers, $\left( n-m \right)<\left( n+m \right)$

Splitting 105 in two factors, we get

$\Rightarrow \left( n-m \right)\left( n+m \right)=1\times 105$

For $\left( n-m \right)=1$ and $\left( n+m \right)=105$, $(m,n)=(52,53)$

$\Rightarrow \left( n-m \right)\left( n+m \right)=3\times 35$

For $\left( n-m \right)=3$ and $\left( n+m \right)=35$, $(m,n)=(16,19)$

$\Rightarrow \left( n-m \right)\left( n+m \right)=5\times 21$

For $\left( n-m \right)=5$ and $\left( n+m \right)=21$, $(m,n)=(8,13)$

$\Rightarrow \left( n-m \right)\left( n+m \right)=7\times 21$

For $\left( n-m \right)=7$ and $\left( n+m \right)=21$, $(m,n)=(4,11)$

Hence there are four pairs.

Let the work be of 40 units

Amount of work done by Anil in one day = 40/20=2 units

Amount of work done by Sunil in one day = 20/20=1 units

Bimal does 10% work i.e. 4 units.

Rest 40-4=36 units is done by Anil and Sunil.

Let Anil took x days. Therefore, Sunil took (x-3) days. Therefore,

$2\times x+1\times \left( x-3 \right)=36$

Or x = 13 days.

Refer to the figure

SO=4-r.

Applying Pythagoras theorem in triangle POS, we get

${{\left( 4+r \right)}^{2}}={{4}^{2}}+{{\left( 4-r \right)}^{2}}$

$\Rightarrow {{\left( 4+r \right)}^{2}}-{{\left( 4-r \right)}^{2}}=16$

$\Rightarrow 4\times 4\times r=16$

$\Rightarrow r=1$

Given A= 72

Also, A=0.9×B => B=A/0.9=72/0.9=80.

And B=1.25×C => C = B/1.25=80/1.25=64

And C=0.8×D => D =C/0.8 = 64/0.8=80.

Time taken by cyclist to cover the distance AB = 60 min

Given, starting from 10:01 am, every minute a motor cycle leaves A and moves towards B. Forty-five such motor cycles reach B by 11 am.

Also, the speed of all the motor cycles is same.

That means that the 45th moto cyclevwhich started at 10:45 am, reached B exactly at 11 am. Rest all reached B some time before B.

Therefore, each motor cycle takes 15 min to cover the distance AB.

Now, if the cyclist doubles his speed, then he will reach B in 30 min i.e. at 10:30 am.

So, the 15th motor cycle (started at 10:15 am from A) would be the last motor cycle to reach point B at 10:30 am.

Hence, there will be 15 motor cycles would have reached B by the time the cyclist reached B.

Let a be the average of 20 numbers whose average does not exceed 5.

Let b be the average of rest of the 10 numbers. Clearly, b>5 i.e. the average of these numbers exceeds 5.

Therefore,

$30\times 5=20a+10b$

$\Rightarrow 2a+b=15$

$\Rightarrow b=15-2a$

Going by the options, we can say that when a=4.5, b=6 which satisfies all the conditions.

Speed of John = 6kmph = $6\times \frac{5}{18}=\frac{5}{3}m/s$

Speed of Mary = 7.5 kmph = $7.5\times \frac{5}{18}=\frac{25}{12}m/s$

Let the track length of A and B be x and y respectively.

Given, x + y = 325 …. (1)

Time taken by John to cover one round of A = $\frac{x}{5/3}\sec $

Therefore, time taken to cover 9 rounds = $9\times \frac{x}{5/3}=\frac{27}{5}x\ \sec $

Time taken by Mary to cover one round of B = $\frac{y}{25/12}\sec $

Therefore, time taken to cover 5 rounds = $5\times \frac{y}{25/12}=\frac{12}{5}y\ \sec $

As per the condition:

$\frac{27}{5}x=\frac{12}{5}y$

$\Rightarrow \frac{x}{y}=\frac{12}{27}=\frac{4}{9}$

Putting in equation (1) we get x=100 and y =225.

Time taken by Mary to cover one round of A=$\frac{100}{25/12}=48\ \sec $

Let the number be ABCDEF, where A, B, C, D, E, and F be the digits.

Given,

C=A

B=2A

F=A+B+C=A+2A+A=4A

E=A+B=A+2A=3A

D=E+F=3A + 4A=7A.

Since A and D both are digit, the maximum possible value of A=1. Therefore, the maximum value of D=7.

Sum of roots = 4a+3a=7a=-b

Or b=-7a

Product of roots = 4a×3a =c

Or $c=12{{a}^{2}}$

Now, ${{b}^{2}}+c={{(-7a)}^{2}}+12{{a}^{2}}=61{{a}^{2}}$

Comparing the options.

Option 1: $61{{a}^{2}}=3721$ $\Rightarrow {{a}^{2}}=61$, clearly a is not an integer.

Option 2: $61{{a}^{2}}=549$ $\Rightarrow {{a}^{2}}=9$, we can have a =-3 or 3 (an integer)

Option 3: $61{{a}^{2}}=427$ $\Rightarrow {{a}^{2}}=7$, clearly a is not an integer.

Option 4: $61{{a}^{2}}=361$ $\Rightarrow {{a}^{2}}=\frac{361}{61}$, clearly a is not an integer.

$2^{6 x}+2^{3 x} \times 2^{2}-21=0$

Take $2^{3 x}=y$

$\Rightarrow y^{2}+4 y-21=0$

$\Rightarrow(y-3)(y+7)=0$

$\Rightarrow y=3$ or $y=-7$

$\Rightarrow 2^{3 x}=3$ or $2^{3 x}=-7\{\text { No solution }\}$

$\Rightarrow 3 x=\log _{2}{3}$

$\Rightarrow x=\frac{log _{2}{3}}{3}$

for $n=1, \quad a_{1}=n \Rightarrow a_{1}=1$

for $n=2, \quad a_{1}-a_{2}=2 \Rightarrow a_{2}=-1$

for $n=3, \quad a_{1}-a_{2}+a_{3}=3 \Rightarrow a_{3}=1$

for $n=4, \quad a_{1}-a_{2}+a_{3}-a_{4}=4 \Rightarrow a_{4}=-1$

From the pattern, each odd term = 1 and each even term = -1

$\Rightarrow a_{51}+a_{52}+\cdots+a_{1022}=0$

Therefore the value is equal to $a_{1023}=1$

Let the track length be 10x.

When they meet at 10 am, ant A travelled 6x of the distance and ant B travelled 4x of the distance.

Therefore, $\frac{\text{Speed of ant A}}{\text{Speed of ant A}}=\frac{6x}{4x}=\frac{3}{2}$

And, the ratio of time taken by A and B to cover the same distance = $\frac{2}{3}$

The distance by ant A from meeting point to point P was 4x. Similarly, the distance covered by ant B from meeting point to point P was 6x.

Given, ant A took 12 min to reach P.

Therefore, to cover a distance of 4x, time taken by ant B = $\frac{3}{2}\times 12=18$min.

But, ant B has to cover a total of 6x distance.

Hence, the time required =$\frac{6x}{4x}\times 18=27$ min.

Therefore, ant B reaches P at 10:27 am.

Refer to the figure below:

Draw the third median CF. We know the following facts:

1. The intersection point of medians i.e. centroid (G) divides each median into 2:1.

2. All three medians divide the triangle into 6 parts of equal area.

$GD=\frac{1}{3}\times AD=\frac{1}{3}\times 12=4$

$GB=\frac{2}{3}\times BE=\frac{2}{3}\times 9=6$

Area of triangle BGD = $\frac{1}{2}\times GB\times GD=\frac{1}{2}\times 6\times 4=12$

Hence, area of triangle ABC = $6\times 12=72$

Let there number of fiction and non fiction books in 2010 be x and y respectively.

From the first condition:

$x+y=11500 \cdots(1)$

From the second condition:

$1.1 \times x+1.2 \times y=11500 \cdots(2)$

Solving both the equations, we get x = 6000.

In 2015, the number of fiction books = 1.1x = 6600

Initial amount of salt in vessel A=10 gms per 100 ml, therefore in 500 ml the amount of salt =50 gms

Initial amount of salt in vessel B=22 gms per 100 ml, therefore in 500 ml the amount of salt =110 gms

Initial amount of salt in vessel C=32 gms per 100 ml, therefore in 500 ml the amount of salt =160 gms

When 100 ml is trasfered from A to B, the amount of salt now in B = 10+110=120 gms in 600ml.

The new concentration of salt in B = 120/600=20 gms per 100 ml.

Also, the amount of salt lef in A =50-10 =40 gms in 400ml.

Now, when 100 ml is trasfered from B to C, the amount of salt now in C = 20+160=180 gms in 600ml.

The new concentration of salt in C= 180/600=30 gms per 100 ml.

Finally, when 100 ml is trasfered from C to A, the amount of salt now in A = 30+40=70 gms in 500ml.

Therefore, the strength of salt in A = $\frac{70}{500}\times 100=14%$

${{2}^{4}}\times {{3}^{5}}\times {{10}^{4}}={{2}^{8}}\times {{3}^{5}}\times {{5}^{4}}$

For perfect squares, we have to take only even powers of the prime factors of the number.

The number of ways 2’s can be used is 5 i.e. ${{2}^{0}},\ {{2}^{2}},\ {{2}^{4}},\ {{2}^{6}},\ {{2}^{8}}\ $

The number of ways 3’s can be used is 3 i.e. ${{3}^{0}},\ {{3}^{2}},\ {{3}^{4}}\ $

The number of ways 5’s can be used is 3 i.e. ${{5}^{0}},\ {{5}^{2}},\ {{5}^{4}}\ $

Therefore, the total number of factors which are perfect squares = $5\times 3\times 3=45$

But this also includes the number 1. Hence excluding 1, the required number is 45-1=44.

$\frac{{{n}^{2}}+7n+12}{{{n}^{2}}-n-12}=\frac{\left( n+3 \right)\left( n+4 \right)}{\left( n-4 \right)\left( n+3 \right)}=\frac{\left( n+4 \right)}{\left( n-4 \right)}$

$\Rightarrow \frac{\left( n+4 \right)}{\left( n-4 \right)}=\frac{\left( n-4+8 \right)}{\left( n-4 \right)}=1+\frac{8}{\left( n-4 \right)}$

The expression is positive integer if $\frac{8}{\left( n-4 \right)}$ is integer.

Or (n-4) must be factor of 8.

For n to be largest, n-4=8

Or n =12

Shortcut:

We can take a=5, b=0, x=13 and y=0 as values which satisfies all three equations.

Hence, $k=ay-bx=5\times 0-0\times 13=0$

For quadratic equation $a x^{2}+b x+c=0$, the roots are real and distinct if $b^{2}-4 a c>0$

Given, $x^{2}-4 x-\log _{2} A=0$

$\therefore(-4)^{2}-4 \times 1 \times\left(-\log _{2} A\right)>0$

$\Rightarrow 16+4 \log _{2} A>0$

$\Rightarrow \log _{2} A>-4$

$\Rightarrow A>2^{-4}$

$\Rightarrow A>\frac{1}{16}$

Cost of table for Aman = 1.2p

Cost of table for Asim = 0.8p

Aman sells to Bimal at 1.3×1.2p=1.56p =cost of table for Bimal = x

Asim sells table to Barun at 0.7×0.8p = 0.56p = cost of table for Barun = y

Therefore, $\frac{x-y}{p}=\frac{1.56p-0.56p}{p}=1$

Let he cost of each bicycle be x.

From the given condition:

$10x\text{ }+\text{ }2000\text{ }=\text{ }6\times 1.25x\text{ }+\text{ }4\times 0.75x$

$\Rightarrow x=4000$

Let the number of hours for regular and overtime work be x and y respectively.

We have two equations:

$x\text{ }+\text{ }y\text{ }=172...(1)$

$\text{114}y\text{ }=\frac{15}{100}\times 57x...(2)$

On solving both the equations, we get x = 160 and y 12.

Hence, his overtime work =12 hours

To get the minimum number of cylinders, the volume of each of the cylinder must be HCF of 405,783, and 351

$\Rightarrow \text{HCF}\ (405,\ 783,\ 351)=27$

Therefore, number of cylinders of iron $=\frac{405}{27}=15$

and, number of cylinders of aluminum $=\frac{783}{27}=29$

and, number of cylinders of copper $=\frac{351}{27}=13$

Hence, the total number of a cylinders $=15+29+13=57$

Also, volume of each cylinder =27 cc

$\Rightarrow \pi r^{2} h=27$

$\Rightarrow \quad \pi \times 3^{2} \times h=27$

$\Rightarrow \quad h=\frac{3}{\pi}$

And total surface area of each cylinder $=2 \pi r(r+h)$

$=2 \pi \times 3\left(3+\frac{3}{\pi}\right)=18(\pi+1)$

Hence, total surface area of 57 cylinders $=57 \times 18(\pi+1)$

$=1026(\pi+1)$

Refer to the figure:

For this right angle triangle, we have the following relations

${{a}^{2}}+{{b}^{2}}={{20}^{2}}=400....(1)$ and

$AP=\frac{ab}{20}....(2)$

For maximum value of AP, we have to maximize the product ab.

Applying $AM\ge GM$ inequality we get

$\frac{{{a}^{2}}+{{b}^{2}}}{2}\ge \sqrt{{{a}^{2}}\times {{b}^{2}}}$

$\Rightarrow \frac{400}{2}\ge ab$

$\Rightarrow ab\le 200$

Hence the maximum value of ab =200.

Therefore, the maximum value of AP =$\frac{200}{20}=10$

From the diagram, it is obvious that AB is the height of the equilateral triangle and is also the slant height of the pyramid.

Therefore, $AB=\frac{\sqrt{3}}{2}\times side=\frac{\sqrt{3}}{2}\times 20=10\sqrt{3}$

And $AO=\frac{1}{2}\times side=\frac{1}{2}\times 20=10$

Applying Pythagoras theorem in triangle AOB

$O{{B}^{2}}=A{{B}^{2}}-O{{A}^{2}}$

$={{\left( 10\sqrt{3} \right)}^{2}}-{{10}^{2}}$

$=200$

Hence, the height of the pyramid (OB) =$10\sqrt{2}$

Given, f(mn)=f(m)f(m)

Also, f(24)=54

$\Rightarrow $f(24) =2×3×3×3

$\Rightarrow $f(2×12)=f(2)f(12)=f(2)f(2×6)=f(2)f(2)f(6)=f(2)f(2)f(2×3)=f(2)f(2)f(2)f(3)=2×3×3×3

Given that f(1), f(2) and f(3) are all positive integers, by comparison, we get

f(2) = 3 and f(3)=2. And we can safely take f(1)=1

Now, f(18)=f(2)(9)=f(2)f(3×3)=f(2)f(3)f(3)=3×2×2=12

The sequence $(2 n+1)+(2 n+3)+(2 n+5)+\ldots+(2 n+47)=5280,$ is in arithmetic progression with first term (a) = 2n+1, common difference (d) = 2 and last term ($t_{n}$)=2n+47.

Let ‘m’ be the number of terms in this sequence.

The last term of A.P. is given by a+(n-1)d

$\Rightarrow \left( 2n+1 \right)+(m-1)(2)=2n+47$

$\Rightarrow m=24$

Also,

$(2 n+1)+(2 n+3)+(2 n+5)+\ldots+(2 n+47)=5280,$

$=\frac{24}{2}\left[ 2\left( 2n+1 \right)+\left( 24-1 \right)\times 2 \right]$

$=24\left( 2n+1+23 \right)=48\left( n+12 \right)$

Therefore, $48\left( n+12 \right)=5280\Rightarrow n=98$

Hence, $1+2+3+\ldots +n=\frac{n(n+1)}{2}=\frac{98\times 99}{2}=4851$

Taking $2^{n d}$ equation

$5^{x-1}+3^{y+1}=9686$, the last digit of $5^{x-1}$ will always be 5 for all positive integral values of x

The power cycle of 3 is:

$3^{4 k+1} \equiv 3$

$3^{4 k+2} \equiv 9$

$3^{4 k+3} \equiv 7$

$3^{4 k} \equiv 1$

Clearly $3^{y+1}$ must be in the form of $3^{4 k}$ as the unit digit of R.H.S. =6

We have ${{3}^{4}}=81,\text{and}\ {{3}^{8}}=6561$

Also, $9686-81=9605$and $9686-6561=3125$

Observe that $3125=5^{5}$

Hence $5^{x-1}=5^{5}$

or $x=6$ and $3^{y+1}=3^{8} \Rightarrow y=7$

(x=6 and y=7 also satisfies the first equation)

Therefore, $x+y=6+7=13$

The formula for each interior angle =$180-\frac{360}{n}$ , where ‘n’ is the side of the regular polygon

$\Rightarrow 180-\frac{360}{2 a}=\frac{3}{2}\left(180-\frac{360}{a}\right)$

$\Rightarrow 360-\frac{360}{a}=540-\frac{3 \times 360}{a}$

$\Rightarrow \frac{2 \times 360}{a}=180$

$\Rightarrow a=\frac{2 \times 360}{180}$

$\Rightarrow a=4$ and $b=2 a=8$

Polygon with each side =$a+b=4+8=12$, will have each interior angle $=180- \frac{360}{12}$

=150

Both the sequences are in arithmetic progression.

The common difference (${{d}_{1}}$ ) for the first sequence = 4

The common difference (${{d}_{2}}$ ) for the first sequence = 5

The first term common is 19.

The common terms will also be in arithmetic progression with common difference $LCM\left( {{d}_{1}},{{d}_{2}} \right)=LCM\left( 4,5 \right)=20$

Let there be ‘n’ terms in this sequence, then the last term would be ≤ 415

i.e. $a+\left( n-1 \right)d\le 415$

$\Rightarrow 19+\left( n-1 \right)\times 20\le 415$

$\Rightarrow \left( n-1 \right)\times 20\le 415-19$

$\Rightarrow \left( n-1 \right)\times 20\le 396$

$\Rightarrow \left( n-1 \right)=\left[ \frac{396}{20} \right]$ where [ ] is the greatest integer

$\Rightarrow \left( n-1 \right)=19$

$\Rightarrow n=20$

Let the amount invested by Bimal be Rs. P

Given, the interest incomes for both are equal. Therefore,

$\left[ 12000\left( 1+\frac{8}{100} \right)-12000 \right]+\left[ 10000{{\left( 1+\frac{3}{100} \right)}^{2}}-10000 \right]=\frac{P\times 7.5\times 1}{100}$

Solving for P we get P =Rs. 20920